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Unformatted text preview: one, E 111 , then in error? Answer: The energies are E x 1 = ¯ h 2 π 2 2 mℓ 2 x E y 1 = ¯ h 2 π 2 2 mℓ 2 y E z 1 = ¯ h 2 π 2 2 mℓ 2 z . If ℓ y and ℓ z are ten times smaller than ℓ x then E y 1 and E z 1 are each 100 times larger than E x 1 . So the onedimensional ground state energy E x 1 is smaller than the true ground state energy E 111 = E x 1 + E y 1 + E z 1 by a factor 201. Which means it is o² by 20,000%. 3.5.8.2 Solution pipegb Question: At what ratio of ℓ y /ℓ x does the energy E 121 become higher than the energy E 311 ? Answer: Using the given expression for E n x n y n z , E n x n y n z = n 2 x ¯ h 2 π 2 2 mℓ 2 x + n 2 y ¯ h 2 π 2 2 mℓ 2 y + n 2 z ¯ h 2 π 2 2 mℓ 2 z , E 121 = E 311 when ¯ h 2 π 2 2 mℓ 2 x + 4¯ h 2 π 2 2 mℓ 2 y + ¯ h 2 π 2 2 mℓ 2 z = 9¯ h 2 π 2 2 mℓ 2 x + ¯ h 2 π 2 2 mℓ 2 y + ¯ h 2 π 2 2 mℓ 2 z...
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This note was uploaded on 01/06/2012 for the course PHY 3604 taught by Professor Dr.danielarenas during the Fall '11 term at UNF.
 Fall '11
 Dr.DanielArenas
 mechanics

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